Date | May Specimen paper | Marks available | 2 | Reference code | SPM.2.AHL.TZ0.11 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | 11 | Adapted from | N/A |
Question
A large tank initially contains pure water. Water containing salt begins to flow into the tank The solution is kept uniform by stirring and leaves the tank through an outlet at its base. Let x grams represent the amount of salt in the tank and let t minutes represent the time since the salt water began flowing into the tank.
The rate of change of the amount of salt in the tank, dxdt, is described by the differential equation dxdt=10e−t4−xt+1.
Show that t + 1 is an integrating factor for this differential equation.
Hence, by solving this differential equation, show that x(t)=200−40e−t4(t+5)t+1.
Sketch the graph of x versus t for 0 ≤ t ≤ 60 and hence find the maximum amount of salt in the tank and the value of t at which this occurs.
Find the value of t at which the amount of salt in the tank is decreasing most rapidly.
The rate of change of the amount of salt leaving the tank is equal to xt+1.
Find the amount of salt that left the tank during the first 60 minutes.
Markscheme
METHOD 1
I(t)=e∫P(t)dt M1
e∫1t+1dt
= eln(t+1) A1
=t+1 AG
METHOD 2
attempting product rule differentiation on ddt(x(t+1)) M1
ddt(x(t+1))=dxdt(t+1)+x
=(t+1)(dxdt+xt+1) A1
so t+1 is an integrating factor for this differential equation AG
[2 marks]
attempting to multiply through by (t+1) and rearrange to give (M1)
(t+1)dxdt+x=10(t+1)e−t4 A1
ddt(x(t+1))=10(t+1)e−t4
x(t+1)=∫10(t+1)e−t4dt A1
attempting to integrate the RHS by parts M1
=−40(t+1)e−t4+40∫e−t4dt
=−40(t+1)e−t4−160e−t4+C A1
Note: Condone the absence of C.
EITHER
substituting t=0,x=0⇒C=200 M1
x=−40(t+1)e−t4−160e−t4+200t+1 A1
using −40e−t4 as the highest common factor of −40(t+1)e−t4 and −160e−t4 M1
OR
using −40e−t4 as the highest common factor of −40(t+1)e−t4 and −160e−t4 giving
x(t+1)=−40e−t4(t+5)+C (or equivalent) M1A1
substituting t=0,x=0⇒C=200 M1
THEN
x(t)=200−40e−t4(t+5)t+1 AG
[8 marks]
graph starts at the origin and has a local maximum (coordinates not required) A1
sketched for 0 ≤ t ≤ 60 A1
correct concavity for 0 ≤ t ≤ 60 A1
maximum amount of salt is 14.6 (grams) at t = 6.60 (minutes) A1A1
[5 marks]
using an appropriate graph or equation (first or second derivative) M1
amount of salt is decreasing most rapidly at t = 12.9 (minutes) A1
[2 marks]
EITHER
attempting to form an integral representing the amount of salt that left the tank M1
60∫0x(t)t+1dt
60∫0200−40e−t4(t+5)(t+1)2dt A1
OR
attempting to form an integral representing the amount of salt that entered the tank minus the amount of salt in the tank at t = 60(minutes)
amount of salt that left the tank is 60∫010e−t4dt−x(60) A1
THEN
= 36.7 (grams) A2
[4 marks]